Since it is so simple
to find the y-intercept
(and it will probably be a point in my T-chart anyway), they are only
asking for the x-intercepts
this time. To find the x-intercept,
I set y
equal 0
and solve:

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0 = –x2
– 4x + 2

x2
+ 4x – 2 = 0

For graphing purposes,
the intercepts are at about (–4.4,
0) and (0.4,
0). (When I write
down the answer, I will of course use the "exact" form, with
the square roots; my calculator's decimal approximations are just for
helping me graph.)

To find the vertex, I
look at the coefficients: a
= –1 and b
= –4. Then:

h =
–(–4)/2(–1) = –2

To find k,
I plug h
= –2 in for x
in y
= –x2
– 4x + 2, and
simplify:

k
= –(–2)2 – 4(–2) + 2 = –4 + 8 + 2 = 10 – 4 = 6

Now I'll find some additional
plot points, to help me fill in my graph:

Note that I picked x-values
that were centered around the x-coordinate
of the vertex. Now I'll plot the parabola:

The vertex is
at (–2, 6),
and the intercepts are at
the following points:

(0,
2), ,
and

Find the x-intercepts
and vertex of y
= –x2 + 2x – 4.

To find the vertex, I
look at the coefficients: a
= –1 and b
= 2. Then:

h
= –(2)/2(–1) = 1

To find k,
I'll plug h
in for x
and simplify:

k
= –(1)2 + 2(1) – 4 = –1 + 2 – 4 = 2 – 5 = –3

The vertex is below the
x-axis,
and, since this is a negative quadratic, I know that the parabola is
going to be upside-down. So can my line possibly cross the x-axis?
Can there possibly be any x-intercepts?
Of course not! So I expect to get "no (real) solution" when
I try to find the x-intercepts,
but I need to show my work anyway. To find the x-intercept,
I set y
equal 0
and solve:

0 = –x2
+ 2x – 4

x2
– 2x + 4 = 0

As soon as I get a negative
inside the square root, I know that I can't get a graphable solution.
So, as expected, there are no x-intercepts.
Now I'll find some additional plot points, to fill in my graph:

This last exercise illustrates
one way you can cut down a bit on your work. If you solve for the vertex
first, then you can easily tell if you need to continue on and look for
the x-intercepts,
or if you can go straight on to plotting some points and drawing the graph.
If the vertex is below the x-axis
(that is, if the y-value
is negative) and the quadratic is negative (so the parabola opens downward),
then there will be no x-intercepts.
Similarly, if the vertex is above the x-axis
(that is, if the y-value
is positive) and the quadratic is positive (so the parabola opens upward),
then there will be no x-intercepts.

In most of the graphs that
I did (though not the first one), it just so happened that the points
on the T-chart were symmetric about the vertex; that is, that the points
"matched" on either side of the vertex. While a parabola is
always symmetric about the vertical line through the vertex (the parabola's
"axis"), the T-chart points might not be symmetric. In particular,
the T-chart points will not "match" if the x-coordinate
of the vertex is something other than a whole number or a half-number
(such as "3.5").
Warning: Don't expect the plot-points always to "match up" on
either side of the vertex; in particular, don't do half the points on
your T-chart and then "fill in" the rest of your T-chart by
assuming a symmetry that might not exist.

Other tips for graphing:
If the parabola is going to be "skinny", then expect that you
will get some very large values in your T-chart. You will either end up
with a really tall graph or else a rather short T-chart. If the parabola
is going to be "fat", then expect that you will probably have
to plot points with fractions as coordinates. In either case, when you
go to connect the dots to draw the parabola, you might find it helpful
to turn the paper sideways and first draw the really curvy part through
the vertex, making sure that it looks nice and round. Then turn the paper
back right-side-up and draw the "sides" of the parabola.

Warning: Draw your graphs
big enough to be clearly seen by your instructor. If you're fitting more
than two or maybe three graphs on one side of a standard sheet of paper,
then you're drawing your graphs way too small.